Monotony criterion

This article deals with the monotony criterion for sequences and series; for the monotony of functions see real monotonic function .

According to the monotonicity criterion, a monotonically falling, downwardly bounded sequence converges towards a limit value.

The monotony criterion , also the main criterion or criterion of monotonous convergence , is an important convergence criterion for sequences and series in mathematics . With the monotonicity criterion, the convergence of a bounded and monotonically increasing or decreasing sequence of real numbers can be proven without knowing its exact limit . The same applies to series with non-negative or non-positive summands.

Monotony criterion for consequences

criteria

The monotony criterion for consequences is:

A monotonically growing sequence of real numbers converges if and only if and only if it is bounded above (equivalent: the sequence has a limit value ) .

Since the convergence behavior of a sequence does not depend on a finite number of first sequence terms, it is sufficient as a prerequisite that the sequence behaves monotonically from a certain sequence term onwards. So is there an index in a sequence of real numbers such that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ ${\displaystyle N\in \mathbb {N} }$

${\displaystyle a_{n}\leq a_{n+1}}$

is for all , and there is further a real bound such that ${\displaystyle n\geq N}$ ${\displaystyle K}$

${\displaystyle a_{n}\leq K}$

is for all , then this sequence converges and holds for the limit value ${\displaystyle n\geq N}$

${\displaystyle \lim _{n\to \infty }a_{n}\leq K}$.

Analogously, a monotonically falling sequence converges if and only if it is bounded below, and its limit value is then at least as large as the lower bound. With the monotony criterion, the existence of the limit value of a monotonic sequence can thus be proven without the exact limit value being known.

proof

The case of a monotonically growing and upwardly restricted sequence is considered . ${\displaystyle (a_{n})}$

Step A.

First it is shown that an upwardly bounded sequence which increases monotonically for almost all terms is convergent.

By assumption, the set has almost all sequence members

${\displaystyle A=\{a_{n},n\geq N\}}$

a supremum because it is limited. ${\displaystyle a=\sup A}$

Be chosen at will. Since there is no smaller upper bound than , there is no upper bound from the index onwards . Therefore applies ${\displaystyle \varepsilon >0}$ ${\displaystyle A}$ ${\displaystyle a=\sup A}$ ${\displaystyle a-\varepsilon }$ ${\displaystyle N}$ ${\displaystyle (a_{n})}$

${\displaystyle a-\varepsilon <a_{M}\leq a}$.

for an appropriately chosen index . Since the index increases monotonically, the following applies ${\displaystyle M\geq N}$ ${\displaystyle (a_{n})}$ ${\displaystyle N}$

${\displaystyle a-\varepsilon <a_{m}\leq a}$

for everyone . So is ${\displaystyle m>M\geq N}$

${\displaystyle |a_{m}-a|=a-a_{m}<\varepsilon }$,

and thus the sequence converges (against the supremum of almost all of its members). ${\displaystyle a}$

Step B.

It remains to be shown that a convergent sequence that grows monotonically for almost all terms is bounded above. The evidence is indirect .

${\displaystyle a=\lim _{n\to \infty }a_{n}}$

be the limit of a sequence that increases monotonically from the index . The existence of a sequence member is assumed ${\displaystyle N}$

${\displaystyle a_{n_{0}\geq N}>a}$.

Since is monotonically increasing for almost all , the following applies ${\displaystyle (a_{n})}$ ${\displaystyle a_{n}}$

${\displaystyle a_{n}\geq a_{n_{0}}\qquad }$ (1)

for everyone . ${\displaystyle n>n_{0}\geq N}$

Be chosen. Then there is one such that applies to everyone : ${\displaystyle \varepsilon =a_{n_{0}}-a>0}$ ${\displaystyle M\geq n_{0}}$ ${\displaystyle m>M\geq n_{0}}$

${\displaystyle a_{m}<a+\varepsilon =a_{n_{0}}}$,

contrary to (1). So does not exist , and is limited upwards for everyone by their limit value . ${\displaystyle a_{n_{0}\geq N}>a}$ ${\displaystyle (a_{n})}$ ${\displaystyle n\geq N}$ ${\displaystyle a}$

${\displaystyle \Box }$

It can be shown quite analogously that:

• a monotonically decreasing, downwardly bounded sequence (towards the infimum of almost all of its terms) converges, and that
• a monotonically falling, convergent sequence is bounded below by its limit value.

example

The consequence with the rule

${\displaystyle a_{n}={\frac {n}{n+1}}}$

is growing monotonously, there

${\displaystyle a_{n}={\frac {n}{n+1}}={\frac {n(n+2)}{(n+1)(n+2)}}<{\frac {n(n+2)+1}{(n+1)(n+2)}}={\frac {(n+1)^{2}}{(n+1)(n+2)}}={\frac {n+1}{n+2}}=a_{n+1}}$,

and it applies

${\displaystyle a_{n}={\frac {n}{n+1}}={\frac {n+1-1}{n+1}}=1-{\frac {1}{n+1}}<1}$

for everyone . Thus the sequence converges to a limit value with ${\displaystyle n}$

${\displaystyle \lim _{n\to \infty }a_{n}\leq 1}$.

As you can see in this example, the limit value of a sequence can be equal to the specified limit, even if each term in the sequence is really smaller than the limit.

application

In practice, the monotony criterion is often used in such a way that for a monotonically growing sequence one finds a monotonically falling sequence that satisfies for all . Then converge both as well and it is ${\displaystyle (a_{n})}$ ${\displaystyle (b_{n})}$ ${\displaystyle a_{n}\leq b_{n}}$ ${\displaystyle n\geq N}$ ${\displaystyle (a_{n})}$ ${\displaystyle (b_{n})}$

${\displaystyle \lim _{n\to \infty }a_{n}\leq \lim _{n\to \infty }b_{n}}$.

For example, the sequence used to define Euler's number is

${\displaystyle a_{n}=\left(1+{\frac {1}{n}}\right)^{n}}$

monotonously growing and the consequence

${\displaystyle b_{n}=\left(1+{\frac {1}{n-1}}\right)^{n}}$

falling monotonously. After it holds, both sequences converge. If (as in this example) a null sequence is formed , then interval nesting is present and it even applies ${\displaystyle a_{n}<b_{n}}$ ${\displaystyle b_{n}-a_{n}}$

${\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}}$.

Monotony criterion for rows

criteria

The monotony criterion for series is:

A series with non-negative real summands converges to a limit if and only if its partial sums are bounded above.

It is also sufficient that the summands are nonnegative above a certain index. So it applies to the summands of a series ${\displaystyle \textstyle \sum _{i=1}^{\infty }a_{i}}$

${\displaystyle a_{i}\geq 0}$

for all and is the consequence of the partial sums ${\displaystyle i\geq N}$ ${\displaystyle (s_{n})_{n\in \mathbb {N} }}$

${\displaystyle s_{n}=\sum _{i=1}^{n}a_{i}\leq K}$

bounded above by a real bound , then this series converges and it holds for the limit value ${\displaystyle K}$

${\displaystyle \sum _{i=1}^{\infty }a_{i}=\lim _{n\to \infty }s_{n}\leq K}$.

Analogously, a series with non-positive real summands converges if and only if their partial sums are bounded below. A series that meets the monotony criterion is not only convergent, but even absolutely convergent .

example

It will be your turn

${\displaystyle \sum _{i=1}^{\infty }{\frac {1}{i^{2}}}=1+{\frac {1}{4}}+{\frac {1}{9}}+{\frac {1}{16}}+\ldots }$

examined for convergence. The summands are all non-negative, so the monotony criterion is applicable. The partial sums of the series are bounded upwards because the inequality applies

${\displaystyle {\frac {1}{i-1}}-{\frac {1}{i}}={\frac {i-(i-1)}{i(i-1)}}={\frac {1}{i^{2}-i}}>{\frac {1}{i^{2}}}}$

and after resolving the resulting telescope sum, the estimate

${\displaystyle s_{n}=\sum _{i=1}^{n}{\frac {1}{i^{2}}}<1+\sum _{i=2}^{n}\left({\frac {1}{i-1}}-{\frac {1}{i}}\right)=1+\left(1-{\frac {1}{n}}\right)<2}$.

Accordingly, the series converges to a limit value which is at most . The actual limit of this series is at . ${\displaystyle 2}$ ${\displaystyle {\tfrac {\pi ^{2}}{6}}=1{,}6449\ldots }$

proof

Here, too, it is sufficient to consider the case of a series with non-negative summands. A series converges when the sequence of its partial sums converges. Off for now follows ${\displaystyle a_{i}\geq 0}$ ${\displaystyle i\geq N}$

${\displaystyle s_{n+1}=s_{n}+a_{n+1}\geq s_{n}}$

for , whereby the sequence of the partial sums increases monotonically from this index onwards. Furthermore, the sequence of the partial sums is subject to an upper limit. The convergence of the partial sum sequence and thus the convergence of the series then follows from the monotony criterion for sequences. ${\displaystyle n\geq N}$ ${\displaystyle (s_{n})}$

See also

• Cauchy criterion

literature

• Harro Heuser : Textbook of Analysis - Part 1 . Springer, 2009, ISBN 3-8348-0777-X .
• Wolfgang Walter : Analysis 1, Volume 1 . Springer, 2004, ISBN 3-540-20388-5 .

Web links

Wikibooks: Math for non-freaks: Monotony criterion for consequences  - learning and teaching materials

Remarks

1. ↑ More information on the concept of the supremum and the existence of the supremum for limited subsets of real numbers can be found in the sections of the article Infimum and Supremum linked here .